Question

A metal ball of radius 0.5 metre at 627 degree celsius is surrounded by a vessel at 27 degree celsius the energy of metal is 0.25

Answer 1

Answer:

From stefan's law we know that,rate of loss of heat is given as,

t

Q

=σA(T

1

4

−T

2

4

)×e

so,T

1

=273+527K=800K

T

2

=273+27K=300K

and A=200×10

−4

m

2

so,

t

Q

=5.67×10

−8

×2×10

−2

[(800

4

)−(300

4

)]×(0.4)=(182.12) Joule